# How do you find the critical points for y=x-sqrtx?

Dec 19, 2016

A critical number is a number in the domain at which the derivative is $0$ or fails to exist.

#### Explanation:

For $f \left(x\right) = x - \sqrt{x}$, the domain is $\left[0 , \infty\right)$

$f ' \left(x\right) = 1 - \frac{1}{2 \sqrt{x}} = \frac{2 \sqrt{x} - 1}{2 \sqrt{x}}$

$f ' \left(x\right) = 0$ $\text{ }$ $\text{ }$ $\text{ }$ $f ' \left(x\right)$ does not exist

$2 \sqrt{x} - 1 = 0$ $\text{ }$ $\text{ }$ $\text{ }$ $2 \sqrt{x} = 0$

$\sqrt{x} = \frac{1}{2}$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$ $x = 0$

$x = \frac{1}{4}$

Both $0$ and $\frac{1}{4}$ are in the domain so both are critical numbers.